HI, If you are using certain amount of given fertilizer for your experiment then you must have added certain dose lets say x% and then you might have use different doses for other treatment like 75% of x, 50% of x and so on. If this is the case you can easily calculate required amount of dose for your given field area

Just convert the required area into m2 (suppose for one acre you got 4,000m2). At the same time convert the experimental plot or pot into m2.  Then figure out the most responsive fertilizer % which you want to use in real field.  Lets say if area is 1m2  for the experiment unit and you get good response when you used 200gm of fertilizer ( suppose having concentration of (10%, 10%,10% NPK).

You can get your required amount 200 gm *4,000m2.

Hope this will help. But I am not entirely sure how did you use the fertilizer for your response experiment

Dear sir you can convert % of fertilizer to ton  / acer as follow:

1- % means one unit of fertilizer per 100 unit soil for example 20% fertilizer means 20 kg fertilizer  per 100 g soil , approximately weight of 1 acer soil = 1.62 million kg soil for 30 cm depth.

2- If amount of applied fertilizer = 2% 

it means adding 2 ton of fertilizer to 100 ton soil or 100000 kg soil .

There for for one acer = 2 * (1.6 million /100000) =32 ton per acer.

the above calculation depends of application of organic fertilizer if the required amount is 2% of soil weight .I am ready to assistance you if you have any question about this subject. 

Many thankings for your help dear Professor Dr Akram Othman Esmail, but  according to this relation 1 acer soil = 1.62 million kg soil for 30 cm depth.the pots i used it in my experiment must be 30 cm2 depth? i mean that the pot depth must take in consedration?

Dear Dr. Hassan,

Prof. Esmail has rightly stated that approximate weight of the plough-layer (30-cm depth) soil of one acre (40,000 sq.m.) field area is 1.62 million kg.

Suppose, you wish to apply your fertilizer at the rate of 200 kg per acre. That means, for 1.62 million kg (i.e. 1620000 kg) soil you need to apply 200 kg fertilizer. That means for 20 kg soil (if taken in each pot), you need to apply (200/1620000)x20 = 0.00247 kg fertilizer (=2.47 g fertilizer). For pot culture study, you need not consider depth of the soil. Only the volume and/or weight of the soil taken in the pot need to be considered for applying fertilizer there. 

Yes, I'm agreed to the answer of Sir Dilip Kumar Kundu.

Moreover, for Dr. Hassan I will like to add about the percent calculation of source of fertilizer which is needed to apply for the given soil. Like if you have to apply Urea to provide N then also consider the percent amount of actual chemical in the fertilizer. Like 46% N in 50 kg Urea bag means it contains 23 kg of actual N in it. So, measure your total amount of fertilizer according to the fertilizer treatment for the specific fertilizer chemical also.

Dear Dr. Hassan,

To know the quantity of your fertilizer to be added per acre, you need two important information:

1. The quantity of soil per acre. Using this information you can easily calculate the amount of your fertilizer required to be added for the soil in your pot based on the quantity of fertilizer required per acre. For example if 100 kg fertilizer is require for 1.62 million kg soil (soil on 1 acre area at the depth of 30 cm), how much fertilizer is required for a soil in your pot (this could be 2 or 3 kg depending on the size of your pot and size and number of your experimental plant/s).

100 kg fertilizer = 1600000 kg soil per acre

X kg fertilizer =  2 or 3 kg soil per pot, then solve for x after cross multiplication. That will be the quantity of fertilizer required per your potting soil.

2. You also need the quantity of nutrients such as N, P and K in kg per 100 kg of your fertilizer. Just like UREA is 46 0 0. This means it contains 46% Nitrogen and 0 % both phosphorus in the form of P2O5 and potassium in the form of K2O. This means 100 kg fertilizer UREA is containing 46 kg nitrogen (N). Then you can easily calculate the quantity of nutrients (N or P or K) per pot by multiplying the quantity of your fertilizer added per your potting soil with the percentage of the Nutrients in your fertilizer material, the fine powders.

recommended rate x 100/%fertilizer

eg if recommended rate is 60kg N per ha, and using 46-0-0 fertilizer

then 60x100/46

A fertilizer company needs to supply 50 tons of fertilizer at the end of the first month, 70 tons at the end of second month, and 90 tons at the end of third month. The cost of producing x tons of fertilizer in any month is given by $(4500x + 20x2). It can produce more fertilizer in any month and supply it in the next month. However, there is an inventory carrying cost of $400 per ton per month. Find the optimal level of production in each of the three periods and the total cost involved by solving it as an initial value problem.

What is the recommended weight of npk needed for tomato in on one hac of land

Two important information are needed. The first is the nutrient content of the new fertilizer, for instance in terms of N, P and K%. The second information needed is recommendation in terms of N, P and K in kg per ha. If you can have these information it is easy to calculate the amount of a fertilizer(fine powders) needed for the nutrition of one acre (can be converted in to ha).

@Dilip Kumar Kundu As I understand fertilizers doses per acre or per ha are calculated by taking soil depth 15cm, not 30 cm. Fertilizer application rates may increase if more weight of soil is taken into consideration. Accordingly weight of 1 acre soil upto 15 centimeters would be 896,000 kgs and one ha 2.24 million kgs.These weights per acre and per ha are being used to calculate fertilizer doses.

If more weight of soil per acre or per ha is taken into consideration and fertlizer nutrient rate per ha is used, in pot culture studies less amount of fertilizer nutrient is calculated per unit weight of soil.

There are numerous factors affecting on amount of fertilizer per hectare:

1- Soil fertility status.

2-Type of plant or crop.

3-Climatic condition such as temperature , amount of rainfall......etc.

4-Chemical composition of irrigation water.

5-Chemical and physical properties of fertilizer or type of fertilizer.

6-Percentage of nutrient or nutrients in fertilizer.